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# User:Jaume Oliver Lafont/A000297

A000297(n)=sum(k=0,n,(k+2)*(2+n-k)) --Jaume Oliver Lafont 10:16, 3 January 2010 (UTC)

${\displaystyle A000297(x)={\frac {(2-x)^{2}}{(1-x)^{4}}}=({\frac {(2-x)}{(1-x)^{2}}})^{2}}$

A000297(n) is the self-convolution of (n+2)*(n>=0) (A-number?)

Other decompositions of the g.f. are:

${\displaystyle {\frac {(2-x)^{2}}{(1-x)^{2}}}*{\frac {1}{(1-x)^{2}}}={\frac {(2-x)^{2}}{(1-x)^{2}}}*A001477(x)/x=(A097950(x)-1-x-x^{2}-2*x^{3}-2*x^{4}-3*x^{5})/x^{6}*A001477(x)/x}$
${\displaystyle {\frac {1}{1-x}}*{\frac {(2-x)^{2}}{(1-x)^{3}}}=A000012(x)*(A034856(x)-x)/x^{2}}$
${\displaystyle {\frac {2-x}{1-x}}*{\frac {(2-x)}{(1-x)^{3}}}=A054977(x)*A000096(x)/x}$
${\displaystyle {\frac {(2-x)^{2}}{1-x}}*{\frac {1}{(1-x)^{3}}}={\frac {(2-x)^{2}}{1-x}}*A000217(x)/x}$

The sequences generated by

${\displaystyle {\frac {(2-x)^{2}}{(1-x)^{2}}}}$
${\displaystyle {\frac {(2-x)^{2}}{1-x}}}$

are

a0: 4,4,5,6,7,8,9,10,11,12,...

a1: 4,0,1,1,1,1,1,1,1,1,...

and a1 is the first difference of a0, which is related to A097950).

--Jaume Oliver Lafont 14:56, 3 January 2010 (UTC)

A000297(n) is square for n=0,2,13,24,27,242,335(...?)

--Jaume Oliver Lafont 20:41, 12 June 2010 (UTC)